Question –
Let $a, b, c$ be non-negative real numbers. Prove that
$$
a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}
$$
my doubt –
author writes
Applying AM-GM inequality, we have
$\sum_{c y c} a^{2} \sqrt{b^{2}-b c+c^{2}}=a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c\right)$
but how we can apply am-gm if there is negative term inside ??? can someone clear this step that how they apply am-gm and got this result..
thankyou
Best Answer
@User88463: For example, look at the term $a^2\sqrt{b^2-bc+c^2}=a\sqrt{a^2(b^2-bc+c^2)}$. Then you simply apply AM-GM to the term $\sqrt{a^2(b^2-bc+c^2)}$: In particular, since you have $2$ non-negative terms, which are $a^2 $ and $b^2-bc+c^2$ respectively, by applying AM-GM to these $2$ terms, you will obtain: $$\frac{a^2+(b^2-bc+c^2)}{2} \ge \sqrt{a^2(b^2-bc+c^2)} \iff \sqrt{a^2(b^2-bc+c^2)} \le \frac{a^2+(b^2-bc+c^2)}{2} $$
Which is simply your AM-GM with $2$ variables. If you still cannot see it, let $A=a^2$ and $B=b^2-bc+c^2$, so in reality you actually have $\frac{A+B}{2} \ge \sqrt{AB}$. Multiplying by $a$ on both sides of your inequality, you will obtain: $$a\sqrt{a^2(b^2-bc+c^2)} \le \frac{a(a^2+b^2+c^2-bc)}{2} $$
Thus: $$\sum_{cyc} a^{2} \sqrt{b^{2}-b c+c^{2}}=\sum_{cyc}a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \le \frac{1}{2} \sum_{cyc} a\left(a^{2}+b^{2}+c^{2}-b c\right)$$